Question: Complete the square to solve for $x$. $x^{2}-16x+63 = 0$
Solution: Begin by moving the constant term to the right side of the equation. $x^2 - 16x = -63$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $-16$ , half of it would be $-8$ , and squaring it gives us ${64}$ $x^2 - 16x { + 64} = -63 { + 64}$ We can now rewrite the left side of the equation as a squared term. $( x - 8 )^2 = 1$ Take the square root of both sides. $x - 8 = \pm1$ Isolate $x$ to find the solution(s). $x = 8\pm1$ So the solutions are: $x = 9 \text{ or } x = 7$ We already found the completed square: $( x - 8 )^2 = 1$